Laplace Transform of sin(t)/t help

[Gunthi]

Homework Statement

\mathcal{L}\left[\frac{\sin(t)}{t}\right]=?

Homework Equations

\mathcal{L}\left[f(t)\right]=\int_0^\infty e^{-zt}f(t)dt

The Attempt at a Solution

\int_0^\infty e^{-zt}\frac{\sin(t)}{t}dt=\int_0^\infty \frac{e^{-zt}}{t}\left(\frac{e^{it}-e^{-it}}{2i}\right )dt

\int_0^\infty f(t)dt=\lim_{T\rightarrow\infty}\int_0^T \frac{1}{t}\left(\frac{e^{t(i-z)}-e^{-t(i+z)}}{2i}\right )

And I end up having to calculate things like
\int_0^T\frac{e^{\alpha t}}{t}dt which doesn't seem to be the best way to do this.

Does anyone know how to do this properly?

Any help would be appreciated, thanks in advance!

 

[snipez90]

First, note that as z tends to infinity, the integral approaches 0 since the kernel e^{-zt} tends to 0 in the limit.

The relevant tool is to differentiate under the integral sign by treating the integral as a function of z. This is justified by Leibniz's rule or more advanced tools from Lebesgue integration (e.g. Lebesgue's dominated convergence theorem). In fact, one approach is to make the substitution for sin as you did, and then differentiate under the integral sign in a distributional sense, and work with Fourier transforms. But simply differentiating under the integral sign with respect to the parameter z from the beginning also suffices.